temperature dependent solubility of thioglycerol-ligated zns

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Temperature Dependent Solubility of Thioglycerol-Ligated ZnS Nanoparticles in 4:1 MeOH:H2O Solution Daniel Scott 1 Dr. Christopher Sorensen 2 Jeff Powell 2 1 Department of Physics, University of Houston 2 Department of Physics, Kansas State University

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Temperature Dependent Solubility of

Thioglycerol-Ligated ZnS Nanoparticles in 4:1 MeOH:H2O Solution

Daniel Scott1

Dr. Christopher Sorensen2

Jeff Powell2

1 Department of Physics, University of Houston2 Department of Physics, Kansas State University

Background

Background

● Significant literature exists for solubility of bulk materials in a wide variety of solvents

Background

● Significant literature exists for solubility of bulk materials in a wide variety of solvents

● Only in the past few decades have nanomaterials become a topic of significant study

Background

● Significant literature exists for solubility of bulk materials in a wide variety of solvents

● Only in the past few decades have nanomaterials become a topic of significant study○ Nanoparticles (NPs) behave differently than bulk

counterparts○ Sparse literature on solubility of NPs

Background

● Treat monodisperse NP colloid in solvent as solution with temperature dependent solubility

Background

● Treat monodisperse NP colloid in solvent as solution with temperature dependent solubility

● Construct equilibrium phase diagram ○ Enthalpy of dissolution

Theory

Theory

● Surface Plasmon Resonance○ Interaction between electrons on surface of NP with

incident light○ Causes unique light absorption profile characteristic

to features such as NP material and size

Theory

● Surface Plasmon Resonance○ Interaction between electrons on surface of NP with

incident light○ Causes unique light absorption profile characteristic

to features such as NP material and size

● UV-Vis spectrometer to view absorption spectrum○ Higher concentration of dissolved NP gives greater

absorption (A=ε*l*c :: Beer-Lambert Law)

Theory

Absorption decreases with lower

concentrations of dissolved NPs

Our System

● ZnS NPs ligated with thioglycerol (3-mercapto-1,2-propanediol)

○ Highly soluble in water

○ Insoluble in methanol

○ SPR peak at ~251nm■ Requires UV-transparent cuvette

Procedure

Procedure

Spectral results

Spectral results

24C

40C

50C

60C

70C

Spectral results

24C

40C

50C

60C

70C

Spectral results

24C

40C

50C

60C

70C

Absorbance decreases at

higher temperatures

Spectral results

24C

40C

50C

60C

70C

Absorbance decreases at

higher temperatures

Less soluble when heated

Exothermic Dissolution

Exothermic Dissolution

In [MeOH + H2O] solution:

ZnS (sc) ⇌ ZnS (c) + heatsc: supercluster (NP aggregates)c: cluster (NP)

Exothermic Dissolution

In [MeOH + H2O] solution:

ZnS (sc) ⇌ ZnS (c) + heatsc: supercluster (NP aggregates)c: cluster (NP)

Equilibrium reaction, thus Le Chatelier’s principle tells us excess heat would favor the left-hand side

Gibbs Free Energy

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

● Exothermic dissolution: ΔHdis < 0, so ΔHfus > 0

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

● Exothermic dissolution: ΔHdis < 0, so ΔHfus > 0● Suppose we increase temperature, forming precipitate:

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

● Exothermic dissolution: ΔHdis < 0, so ΔHfus > 0● Suppose we increase temperature, forming precipitate:

○ ΔHfus - ΔTΔS = (+) - (+) * ΔS

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

● Exothermic dissolution: ΔHdis < 0, so ΔHfus > 0● Suppose we increase temperature, forming precipitate:

○ ΔHfus - ΔTΔS = (+) - (+) * ΔS■ ΔS must be positive for ΔG to be negative so that

precipitation at higher temperatures is spontaneous

Gibbs Free Energy

● Process is spontaneous if ΔG < 0:○ ΔG = ΔH - ΔTΔS

● Exothermic dissolution: ΔHdis < 0, so ΔHfus > 0● Suppose we increase temperature, forming precipitate:

○ ΔHfus - ΔTΔS = (+) - (+) * ΔS■ ΔS must be positive for ΔG to be negative so that

precipitation at higher temperatures is spontaneous

■ Higher entropy (disorder) in precipitate than dissolved

Dissolved: Less Disorder

Dissolved: Less Disorder

3-mercapto-1,2-propanediol ligand (thioglycerol)

Dissolved: Less Disorder

Hydrogen bonding sites

3-mercapto-1,2-propanediol ligand (thioglycerol)

Potential Explanation: Hydrogen Bonds

● Formation of hydrogen bond is highly exothermic

Potential Explanation: Hydrogen Bonds

● Formation of hydrogen bond is highly exothermic○ Hydrogen bonds have a deep potential well

Potential Explanation: Hydrogen Bonds

● Formation of hydrogen bond is highly exothermic○ Hydrogen bonds have a deep potential well○ More energy released in formation of hydrogen bond

than is consumed in destruction of solute-solute (inter-NP) bond

Potential Explanation: Hydrogen Bonds

● Formation of hydrogen bond is highly exothermic○ Hydrogen bonds have a deep potential well○ More energy released in formation of hydrogen bond

than is consumed in destruction of solute-solute (inter-NP) bond

○ Falls to a lower energy state with hydrogen bond

Potential Explanation: Hydrogen Bonds

● Formation of hydrogen bond is highly exothermic○ Hydrogen bonds have a deep potential well○ More energy released in formation of hydrogen bond

than is consumed in destruction of solute-solute (inter-NP) bond

○ Falls to a lower energy state with hydrogen bond

● Dissolved ZnS with hydrogen bonds is more ordered (less disordered) than undissolved as ZnS precipitate

Calculating ΔHdis

Calculating ΔHdis

● By van’t Hoff equation: ln(x) = -(ΔHdis/RT) + c○ x: mole fraction○ R: gas constant (8.314 x 10-3 kJ/mol K)○ T: temperature (Kelvin)○ c: constant related to activity coefficient

Calculating ΔHdis

● By van’t Hoff equation: ln(x) = -(ΔHdis/RT) + c

● Beer-Lambert Law: absorbance (A) proportional to concentration

Calculating ΔHdis

● By van’t Hoff equation: ln(x) = -(ΔHdis/RT) + c

● Beer-Lambert Law: absorbance (A) proportional to concentration

● Colligative property of dilute solutions: concentration approx. proportional to mole fraction

Calculating ΔHdis

● By van’t Hoff equation: ln(x) = -(ΔHdis/RT) + c

● Beer-Lambert Law: absorbance (A) proportional to concentration

● Colligative property of dilute solutions: concentration approx. proportional to mole fraction

● Then x=bA○ ln(bA) = -(ΔHdis/RT) + c

Calculating ΔHdis

● ln(bA) = -(ΔHdis/RT) + c○ Slope of ln(bA) vs (1/T): -(ΔHdis/R)

Calculating ΔHdis

● ln(bA) = -(ΔHdis/RT) + c○ Slope of ln(bA) vs (1/T): -(ΔHdis/R)○ Calculating slope

■ [ln(bA2) - ln(bA1)] / [(1/T2) - (1/T1)]

Calculating ΔHdis

● ln(bA) = -(ΔHdis/RT) + c○ Slope of ln(bA) vs (1/T): -(ΔHdis/R)○ Calculating slope

■ [ln(bA2) - ln(bA1)] / [(1/T2) - (1/T1)]■ [(ln(b) + ln(A2)) - (ln(b) + ln(A1))] / [(1/T2) - (1/T1)]

Calculating ΔHdis

● ln(bA) = -(ΔHdis/RT) + c○ Slope of ln(bA) vs (1/T): -(ΔHdis/R)○ Calculating slope

■ [ln(bA2) - ln(bA1)] / [(1/T2) - (1/T1)]■ [(ln(b) + ln(A2)) - (ln(b) + ln(A1))] / [(1/T2) - (1/T1)]■ [ln(A2) - ln(A1)] / [(1/T2) - (1/T1)]

Calculating ΔHdis

● ln(bA) = -(ΔHdis/RT) + c○ Slope of ln(bA) vs (1/T): -(ΔHdis/R)○ Calculating slope

■ [ln(bA2) - ln(bA1)] / [(1/T2) - (1/T1)]■ [(ln(b) + ln(A2)) - (ln(b) + ln(A1))] / [(1/T2) - (1/T1)]■ [ln(A2) - ln(A1)] / [(1/T2) - (1/T1)]

● Proportionality b does not affect slope

Calculating ΔHdis

Calculating ΔHdis

● Slope 1000/T: m = 4

Calculating ΔHdis

● Slope 1000/T: m = 4○ Slope 1/T: m = 4000

Calculating ΔHdis

● Slope 1000/T: m = 4○ Slope 1/T: m = 4000

● 4000 = -(ΔHdis/R)○ R = 8.314 x 10-3 kJ/mol K

Calculating ΔHdis

● Slope 1000/T: m = 4○ Slope 1/T: m = 4000

● 4000 = -(ΔHdis/R)○ R = 8.314 x 10-3 kJ/mol K

● ΔHdis = -3 x 101 kJ/mol K

Conclusions

● Thioglycerol-ligated ZnS becomes less soluble at higher temperatures in MeOH/H2O solution

Conclusions

● Thioglycerol-ligated ZnS becomes less soluble at higher temperatures in MeOH/H2O solution

● Dissolution is exothermic

Conclusions

● Thioglycerol-ligated ZnS becomes less soluble at higher temperatures in MeOH/H2O solution

● Dissolution is exothermic

● ΔHdis = -3 x 101 kJ/mol K (for 4:1 ratio MeOH:H2O in the region of 40C-70C)

Acknowledgements

For providing the nanoparticles used in this experiment:

Doris Segets, Sebastian SüßFriedrich-Alexander-Universität Erlangen-Nürnberg

For providing the grant funding this REU program:

National Science Foundation

For their mentorship:

Jeff Powell and Dr. Chris Sorensen